EXERCISE 4.3 QUESTION NUMBER 2 CLASS 10TH NCERT MATHS SOLUTION ALL
प्रश्न -(2 ) निम्न द्विघात समीकरण के मूल द्विघाती सूत्र का उपयोग करके ज्ञात कीजिये।
( I ) 2X^2 -7X +3 = 0
(II) 2X^2 + X - 4 =0
(III) 4X^2 + 4\/3 X +3 =0
(IV ) 2X^2 + X +4 =0
हल : ( I ) 2X^2 -7X +3 = 0
दिए गए समीकरण की तुलना a x^2 + b x + c = 0 से करने पर ,
a = 2 , b = -7 , c = 3
एवं , (धन चिन्ह लेने पर )
x = - b + ( \/b^2 - 4 a c) / 2a
x = - (- 7 ) + ( \/ (- 7 )^2 - 4 * 2 * 3 ) / 2 * 2
x = { 7 + (\/ 49 - 24 )} / 4
x = ( 7 + \/25 ) / 4
x = ( 7 + 5 ) / 4
x = 12 / 4
x = 3
( ऋण चिन्ह लेने पर )
x = -b - (\/b^ 2 - 4 a c ) / 2a
x = - (-7 ) - ( \/(-7 )^2 - 4 * 2 * 3 ) / 2 * 2
x = 7 - \/49 -24 / 4
x = (7 - \/25 ) / 4
x = (7 - 5) / 4
x = 2 / 4
x = 1 / 2
उत्तर : अभीष्ट मूल x = 3 , 1 / 2 हे।
(ii) दिया गया द्विघात समीकरण ,
ENGLISH TRANSLATION :
Question -(2 ) Find the following quadratic equation using the original quadratic formula.
( I ) 2X^2 -7X +3 = 0
(II) 2X^2 + X - 4 =0
(III) 4X^2 + 4\/3 X +3 =0
(IV ) 2X^2 + X +4 =0
Solution : ( I ) 2X^2 -7X +3 = 0
Comparing the given equation with a x^2 + b x + c = 0,
a = 2 , b = -7 , c = 3
and , (on taking the plus sign)
x = - b + ( \/b^2 - 4 a c) / 2a
x = - (- 7) + ( \/ (- 7 )^2 - 4 * 2 * 3 ) / 2 * 2
x = { 7 + (\/ 49 - 24 )} / 4
x = ( 7 + \/25 ) / 4
x = ( 7 + 5 ) / 4
x = 12 / 4
x = 3
(taking the minus sign)
x = -b - (\/b^ 2 - 4 a c ) / 2a
x = - (-7 ) - ( \/(-7 )^2 - 4 * 2 * 3 ) / 2 * 2
x = 7 - \/49 -24 / 4
x = (7 - \/25 ) / 4
x = (7 - 5) / 4
x = 2 / 4
x = 1/2
Answer: Required root x = 3 , 1 / 2
(ii) Given quadratic equation,
On taking the plus sign,
On taking the minus sign,
Hence, the roots of the quadratic equation will be ,
(iii ) The given quadratic equation is as follows,